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 Help With Trig Identity		  New Topic  Reply to Topic
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Heather
Junior Member

USA
4 Posts
Posted - 10/23/2007 :  19:37:51  Show Profile  Reply with Quote
I can't figure out this identity, I was wondering if someone could help me with it...

sin-tancos-cot=tan^6
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skeeter
Advanced Member

USA
5634 Posts
Posted - 10/23/2007 :  19:49:14  Show Profile  Reply with Quote
(sinx - tanx)/(cosx - cotx) =

[sinx(1 - secx)]/[cosx(1 - cscx)] =

(sinx)/(cosx)*(1 - secx)/(1 - cscx) =

tanx * (-tanx)/(-cotx) =

tanx * tanx * tanx =

tan6x
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Heather
Junior Member

USA
4 Posts
Posted - 10/23/2007 :  20:16:45  Show Profile  Reply with Quote
Thank you sooo much!! it was like a big DUH when i saw it! haha thanks again!
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